If a, b and c are three non-zero vectors such that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$, then the value of $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}$ will be:

1.	Less than zero	2.	equal to zero
3.	greater than zero	4.	3

If $\stackrel{}{\mathrm{}}$\(\overrightarrow P +\overrightarrow Q\) = \(\overrightarrow P-\overrightarrow Q\) and \(\theta\) is the angle between \(\overrightarrow {P}\) and \(\overrightarrow {Q}\), then:
1. \(\theta = 0^{\circ}\)
2. \(\theta = 90^{\circ}\)
3. \(P=0\)
4. \(Q=0\)

What is the torque of a force $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$newton acting at a point $\overrightarrow{\mathrm{r}}=\left(3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$ metre about the origin? (Given: $\overrightarrow{\mathrm{\tau }}<mspace\; width="1em"></mspace>=\overrightarrow{\mathrm{r}<mspace\; width="1em"></mspace>}\times \overrightarrow{\mathrm{F}}$)

1. $6\hat{\mathrm{i}}-6\hat{\mathrm{j}}+12\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

2. $17\hat{\mathrm{i}}-6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

3. $-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

4. $-17\hat{\mathrm{i}}+6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

The momentum of a body moving in a straight line is $\mathrm{p}=\left({\mathrm{t}}^2+2\mathrm{t}+1\right)<mspace\; width="1em"></mspace>\mathrm{kg}<mspace\; width="1em"></mspace>\mathrm{m}/\mathrm{s}$. Force acting on the body at t=2 sec will be: \(\left(\text{Given:}~ F=\frac{dp}{dt}\right)\)
1. 6 N
2. 8 N
3. 4 N
4. 2 N

The temperature of a body varies with time as \(T= (T_0 +at^2 +b\sin t)K,\) where \(T_0\) is the temperature in Kelvin at \(t =0 ~\text{s}~\text{&}~a = \frac{2}{\pi}~\text{K/s}^2~\text{&}~b = -4~\text{K}\)$\mathrm{}$, then the rate of change of temperature \(\frac{dT}{dt}\) at \(t = \pi ~\text{s}\) is:
1. \(8~\text{K}\)
2. \(80~\text{K}\)
3. \(8~\text{K/sec}\)
4. \(80~\text{K/sec}\)

$\stackrel{}{\mathrm{}}$\(\overrightarrow{A}\) and \(\overrightarrow B\) are two vectors and \(\theta\) is the angle between them. If \(\left|\overrightarrow A\times \overrightarrow B\right|= \sqrt{3}\left(\overrightarrow A\cdot \overrightarrow B\right),\) then the value of \(\theta\) will be:

A particle is moving along the x-axis. The velocity v of this particle varies with its position x as $\mathrm{v}=\frac{1}{\mathrm{x}}$. Find the velocity of the particle as a function of time t given that at t=0, x=1 and v=$\frac{\mathrm{dx}}{\mathrm{dt}}$.

1. $\mathrm{v}=\sqrt{2\mathrm{t}+1}$

2. $\mathrm{v}=\frac{1}{\sqrt{2\mathrm{t}+1}}$

3. $\mathrm{v}=\frac{1}{\mathrm{t}}$

4. None of these

If a curve is governed by the equation y = sinx, then the area enclosed by the curve and x-axis between x = 0 and x =$\mathrm{\pi }$ is (shaded region):
           
1. \(1\) unit
2. \(2\) units
3. \(3\) units
4. \(4\) units

The acceleration of a particle starting from rest varies with time according to relation, $a=\alpha <mspace\; width="1em"></mspace>t+\beta$. The velocity of the particle at time instant \(t\) is: \(\left(\text{Here,}~ a=\frac{dv}{dt}\right)\)

1. $\alpha t^2+\beta t$

2. $\alpha t^2+\frac{\beta t}{2}$

3. $\frac{\alpha t^2}{2}+\beta t$

4. $2\alpha t^2+\beta t$